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A number $+1$ has been entered in 24 fields of the $5 \times 5$ array, and the number $-1$ in one.
In one move, you can multiply by $ -1 $ all numbers in the subarray $ k \times k $ where $ k> 1 $.
Which field at the beginning contained the number $ -1 $, since after a finite number of moves you can get to the state with twentyfive $ + 1 $


I have already proved that that $-1$ cannot be written in fields with $a$ since every subarray contains even number of fields with the letter $a$, therefore the product of the numbers in fields with letter $ a $ would be negative, which leads to a contradiction. \begin{matrix} b & b & a & b & b \\ b & b & a & b & b \\ a & a & 0 & a & a \\ b & b & a & b & b \\ b & b & a & b & b \end{matrix} However I'm not able to to disprove the possibility of $-1$ in fields with letter $b$.
Any help would be greatly appreciated.

  • What about the case that the middle square is $-1$ at the beginning ? – Peter Mar 25 '22 at 08:56
  • Then we can reach the state with twentyfive $+1$ –  Mar 25 '22 at 09:00
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    If you can solve for middle square $(3,3)$ than you can solve for square $(2,2)$ also. Sequence of moves $(1-2;2-3), (2-3,1-2)$, $(1-3,3-5), (3-5,1-3)$, $(1-2,4-5), (4-5,1-2)$, $(3-5,3-5), (4-5,4-5)$ changes sign of exactly $(2,2)$ and $(3,3)$ squares. I use notation $(row_1-row_2,column_1-column_2)$ for moves, where $row_2-row_1=column_2-column_1=k$. – Ivan Kaznacheyeu Mar 25 '22 at 09:44
  • You proved that if you can solve for $(1,2)$, you can solve for $(2,1)$, unless i missed something – caduk Mar 25 '22 at 09:58
  • A tool would be nice where we can make moves and follow the change of the signs. – Peter Mar 25 '22 at 10:06
  • No I'm wrong, but it seems that your sequence leaves (3,3), (1,2), (2,1) from a full $1$ board. – caduk Mar 25 '22 at 10:06

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